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Math Help - Learning Calculus - Finding turning points.

  1. #1
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    Learning Calculus - Finding turning points.

    Hi everyone,

    I'm trying to teach myself calculus and have come across this problem.

    I have a function of x

    f(x) = x^3 + 5x^2 - 8x - 12

    so I set this to zero by

    f'(x) = 3x^2 + 10x - 8 = 0

    then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

    Could someone guide me in the right direction please.

    Thanks in advance

    Andy
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  2. #2
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    Re: Learning Calculus

    Quote Originally Posted by Rastanking View Post
    Hi everyone,

    I'm trying to teach myself calculus and have come across this problem.

    I have a function of x

    f(x) = x^3 + 5x^2 - 8x - 12

    so I set this to zero by

    f'(x) = 3x^2 + 10x - 8 = 0

    then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

    Could someone guide me in the right direction please.

    Thanks in advance

    Andy
    By solving f'(x) = 0 - in this case using your favourite method for solving quadratic equations
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  3. #3
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    Re: Learning Calculus

    Quote Originally Posted by Rastanking View Post
    Hi everyone,

    I'm trying to teach myself calculus and have come across this problem.

    I have a function of x

    f(x) = x^3 + 5x^2 - 8x - 12

    so I set this to zero by

    f'(x) = 3x^2 + 10x - 8 = 0

    then the book I'm using says that I can find the local values of x for min and max and these should be x1 = 0.667 and x2 = -4.000 but it doesn't say how these figures are reached.

    Could someone guide me in the right direction please.

    Thanks in advance

    Andy
    When you have a quadratic equation of the form

    ax^2+bx+c=0

    The "x" solutions are simple to see if you can write your sum of 3 terms as a product by factoring to

    (3x+?)(x+?)=0

    We take the factors of -8 that sum to 10 when multiplying out the brackets.
    This is (3)4+1(-2) and so

    3x^2+10x-8=0\Rightarrow\ (3x-2)(x+4)=0

    Since 0(anything)=0

    then those factors are giving you the 2 solutions

    3x-2=0

    x+4=0

    Alternatively,

    for ax^2+bx+c=0

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
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  4. #4
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    Re: Learning Calculus

    Thanks for the quick reply, it has been a great help. However the answers in the book are 0.667 and -4.000 but I end up with -1.33 and -2.

    I'm either losing the plot entirely or having finger trouble, could someone possible check my maths please.
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  5. #5
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    Re: Learning Calculus

    You should show how you are getting those answers.
    Are you factoring or using the "quadratic formula" ?
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  6. #6
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    Re: Learning Calculus

    I'm using the quadratic formula and ending up with -1.33 and -2
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  7. #7
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    Re: Learning Calculus

    f'(x)=0\Rightarrow\ 3x^2+10x-8=0

    \Rightarrow\ a=3,\;\;b=10,\;\;c=-8

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-10\pm\sqrt{100-4(3)(-8)}}{6}

    x=\frac{-10\pm\sqrt{196}}{6}

    x_1=\frac{\sqrt{196}-10}{6}

    x_2=\frac{-10-\sqrt{196}}{6}

    One of these is positive and the other is negative.

    Your error was to take c=8 instead of minus 8.

    ax^2+bx+c=(3)x^2+(10)x+(-8)
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  8. #8
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    Re: Learning Calculus

    Many thanks again for your help, very much appreciated.
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